I have a convex set $Y \subset \mathbb{R}^n$ such that $0 \notin Y$. I fix $y_1,\dots,y_l \in Y$ I define $C:=C(y_1,\dots,y_l)$ as the strictly convex combinations of these $y_i$'s (strictly meaning that all $y_i$ have weight bigger than $0$).
I already know that $\exists \xi : \langle \xi , c \rangle \ge 0$ for all $c \in C$. Now I want to show that $\exists c \in C$ such that $\langle \xi , c \rangle > 0$.
Two remarks about this: I know that the first inequality works for $\xi$ given by $\Vert \xi \Vert =distance(0,C)$. And also that $\Vert \xi \Vert > 0$ holds. So I can make the first inequality strict and therefore also get the second.
But I want to show the strict inequality without exploiting that $\Vert \xi \Vert > 0$. I want to use the fact that $ c = \lambda_1 y_1 + \dots + \lambda_l y_l$ and at least for one of the $y_i$'s $\Vert y_i \Vert > 0$. Of course, $\Vert y_i \Vert > 0$ for all $i$, but I only one to use the weaker condition. Is this possible?
If we assume that at least for one $y_i$, $|\langle\xi,y_i\rangle| > 0$, then due to continuity of the mapping of $t \in R^{n-1}$, $t \to t_1 \langle\xi,y_1\rangle+\cdots (1-t_1-\cdots t_{n-1})\langle\xi,y_n\rangle$ onto the convex combination there exist an open set around that extreme point for which the inequality holds, pick one suitable $c$ there. Konstructively take $t$ so that $t_i = 1$ for $i$ the coordinate with nonzero $\langle \xi, y_i\rangle$, and zero elsewere. then take $s_j = x/(n-1)$ $j \neq i$ and $s_i=-x$ then $t+s$ is a convex combination and for all small $$ \langle \xi, \sum (t_k+s_k)y_k \rangle = \langle \xi, y_i\rangle + x \left (\left( \frac{1}{n-1}\sum_{k\neq i} \langle \xi, y_k\rangle \right ) - \langle \xi, y_i \rangle \right ) = \langle \xi, y_i\rangle + x C $$ take $x = \min(1/2,\left |\frac{\langle \xi, y_i\rangle}{C}\right |/2)$ if $C$ nonzero else 1/2